Mixing
what does the notation $x^{(n)}$ mean?
$begingroup$
What does the notation $x^{(n)}$ where $x$ is a matrix and $n$ is an integer?
$$|W^Tmathbf{x}^{(n)}+b-y^{(n)}|_2^2$$
optimization notation
edited Jan 16 at 2:01
David M.
1,734418
asked Jan 15 at 20:11
Ibrahim AbouhashishIbrahim Abouhashish
367
$endgroup$
add a comment |
$begingroup$
What does the notation $x^{(n)}$ where $x$ is a matrix and $n$ is an integer?
$$|W^Tmathbf{x}^{(n)}+b-y^{(n)}|_2^2$$
optimization notation
edited Jan 16 at 2:01
David M.
1,734418
asked Jan 15 at 20:11
Ibrahim AbouhashishIbrahim Abouhashish
367
$endgroup$
add a comment |
$begingroup$
What does the notation $x^{(n)}$ where $x$ is a matrix and $n$ is an integer?
$$|W^Tmathbf{x}^{(n)}+b-y^{(n)}|_2^2$$
optimization notation
edited Jan 16 at 2:01
David M.
1,734418
asked Jan 15 at 20:11
Ibrahim AbouhashishIbrahim Abouhashish
367
$endgroup$
What does the notation $x^{(n)}$ where $x$ is a matrix and $n$ is an integer?
$$|W^Tmathbf{x}^{(n)}+b-y^{(n)}|_2^2$$
optimization notation
optimization notation
edited Jan 16 at 2:01
David M.
1,734418
asked Jan 15 at 20:11
Ibrahim AbouhashishIbrahim Abouhashish
367
edited Jan 16 at 2:01
David M.
1,734418
asked Jan 15 at 20:11
Ibrahim AbouhashishIbrahim Abouhashish
367
edited Jan 16 at 2:01
David M.
1,734418
edited Jan 16 at 2:01
David M.
1,734418
edited Jan 16 at 2:01
David M.
1,734418
1,734418
asked Jan 15 at 20:11
Ibrahim AbouhashishIbrahim Abouhashish
367
asked Jan 15 at 20:11
Ibrahim AbouhashishIbrahim Abouhashish
367
asked Jan 15 at 20:11
Ibrahim AbouhashishIbrahim Abouhashish
367
367
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is just an index for a sequence $x^{(1)}, x^{(2)}, x^{(3)}, ldots$, similar to $x_1, x_2, x_3, ldots$.
The reason why they avoid subscripts $x_i$ is because presumably $x$ is a vector, so $x_i$ may be reserved for denoting "the $i$th component of the vector $x$." Using the above notation allows one to do things like $x^{(2)}_3$, which denotes the third component of the vector $x^{(2)}$ (which itself is the second element in a sequence of vectors).
Finally, the reason for the parentheses is to differentiate it from an exponent, since "$y^2$" might look like the square of $y$.
answered Jan 15 at 20:17
angryavianangryavian
41.4k23381
$endgroup$
add a comment |
$begingroup$
As I see the optimization tag, this probably means the matrix that is yielded at the $n$-th iteration step.
In simple words, this means that $x^{(n)}$ is the $x$ matrix of your method at the $n$-th step and $y^{(n)}$ is the $y$ matrix of your method at the $n$-th step.
edited Jan 15 at 20:46
Bernard
121k740116
answered Jan 15 at 20:13
RebellosRebellos
14.8k31248
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
It is just an index for a sequence $x^{(1)}, x^{(2)}, x^{(3)}, ldots$, similar to $x_1, x_2, x_3, ldots$.
The reason why they avoid subscripts $x_i$ is because presumably $x$ is a vector, so $x_i$ may be reserved for denoting "the $i$th component of the vector $x$." Using the above notation allows one to do things like $x^{(2)}_3$, which denotes the third component of the vector $x^{(2)}$ (which itself is the second element in a sequence of vectors).
Finally, the reason for the parentheses is to differentiate it from an exponent, since "$y^2$" might look like the square of $y$.
answered Jan 15 at 20:17
angryavianangryavian
41.4k23381
$endgroup$
add a comment |
$begingroup$
It is just an index for a sequence $x^{(1)}, x^{(2)}, x^{(3)}, ldots$, similar to $x_1, x_2, x_3, ldots$.
The reason why they avoid subscripts $x_i$ is because presumably $x$ is a vector, so $x_i$ may be reserved for denoting "the $i$th component of the vector $x$." Using the above notation allows one to do things like $x^{(2)}_3$, which denotes the third component of the vector $x^{(2)}$ (which itself is the second element in a sequence of vectors).
Finally, the reason for the parentheses is to differentiate it from an exponent, since "$y^2$" might look like the square of $y$.
answered Jan 15 at 20:17
angryavianangryavian
41.4k23381
$endgroup$
add a comment |
$begingroup$
It is just an index for a sequence $x^{(1)}, x^{(2)}, x^{(3)}, ldots$, similar to $x_1, x_2, x_3, ldots$.
The reason why they avoid subscripts $x_i$ is because presumably $x$ is a vector, so $x_i$ may be reserved for denoting "the $i$th component of the vector $x$." Using the above notation allows one to do things like $x^{(2)}_3$, which denotes the third component of the vector $x^{(2)}$ (which itself is the second element in a sequence of vectors).
Finally, the reason for the parentheses is to differentiate it from an exponent, since "$y^2$" might look like the square of $y$.
answered Jan 15 at 20:17
angryavianangryavian
41.4k23381
$endgroup$
It is just an index for a sequence $x^{(1)}, x^{(2)}, x^{(3)}, ldots$, similar to $x_1, x_2, x_3, ldots$.
The reason why they avoid subscripts $x_i$ is because presumably $x$ is a vector, so $x_i$ may be reserved for denoting "the $i$th component of the vector $x$." Using the above notation allows one to do things like $x^{(2)}_3$, which denotes the third component of the vector $x^{(2)}$ (which itself is the second element in a sequence of vectors).
Finally, the reason for the parentheses is to differentiate it from an exponent, since "$y^2$" might look like the square of $y$.
answered Jan 15 at 20:17
angryavianangryavian
41.4k23381
answered Jan 15 at 20:17
angryavianangryavian
41.4k23381
answered Jan 15 at 20:17
angryavianangryavian
41.4k23381
answered Jan 15 at 20:17
angryavianangryavian
41.4k23381
41.4k23381
add a comment |
add a comment |
$begingroup$
As I see the optimization tag, this probably means the matrix that is yielded at the $n$-th iteration step.
In simple words, this means that $x^{(n)}$ is the $x$ matrix of your method at the $n$-th step and $y^{(n)}$ is the $y$ matrix of your method at the $n$-th step.
edited Jan 15 at 20:46
Bernard
121k740116
answered Jan 15 at 20:13
RebellosRebellos
14.8k31248
$endgroup$
add a comment |
$begingroup$
As I see the optimization tag, this probably means the matrix that is yielded at the $n$-th iteration step.
In simple words, this means that $x^{(n)}$ is the $x$ matrix of your method at the $n$-th step and $y^{(n)}$ is the $y$ matrix of your method at the $n$-th step.
edited Jan 15 at 20:46
Bernard
121k740116
answered Jan 15 at 20:13
RebellosRebellos
14.8k31248
$endgroup$
add a comment |
$begingroup$
As I see the optimization tag, this probably means the matrix that is yielded at the $n$-th iteration step.
In simple words, this means that $x^{(n)}$ is the $x$ matrix of your method at the $n$-th step and $y^{(n)}$ is the $y$ matrix of your method at the $n$-th step.
edited Jan 15 at 20:46
Bernard
121k740116
answered Jan 15 at 20:13
RebellosRebellos
14.8k31248
$endgroup$
As I see the optimization tag, this probably means the matrix that is yielded at the $n$-th iteration step.
In simple words, this means that $x^{(n)}$ is the $x$ matrix of your method at the $n$-th step and $y^{(n)}$ is the $y$ matrix of your method at the $n$-th step.
edited Jan 15 at 20:46
Bernard
121k740116
answered Jan 15 at 20:13
RebellosRebellos
14.8k31248
edited Jan 15 at 20:46
Bernard
121k740116
edited Jan 15 at 20:46
Bernard
121k740116
edited Jan 15 at 20:46
Bernard
121k740116
121k740116
answered Jan 15 at 20:13
RebellosRebellos
14.8k31248
answered Jan 15 at 20:13
RebellosRebellos
14.8k31248
answered Jan 15 at 20:13
RebellosRebellos
14.8k31248
14.8k31248
add a comment |
add a comment |
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